Question

A body of mass $m$ is kept on a rough fixed inclined plane of angle of inclination $\theta$. It remains stationary. Then magnitude of force acting on the body by the inclined plane is equal to

• A. $\mu mgcos\theta$
• B. $mgcos\theta$
• C. $mgsin\theta$
• D. $mg$

Answer 1

Answer: (D)

Friction force by the inclined plane is $=\mu mg\cos \theta =mg\sin \theta$ up the inclined plane since the block does not slide.
Normal force perpendicular to inclined plane is $mg\cos \theta$.
So, Total force is vector sum of two forces is equal to $mg$.
$<br/>\begin{array}{l}<br/>N=mg\cos \theta ,f_s=mg\sin \theta \\ <br/>R^2=N^2+f_s^2\Rightarrow R=mg<br/>\end{array}<br/>$